Question

Find `"dy"/"dx"`, if : `x = cos^-1(4t^3 - 3t), y = tan^-1(sqrt(1 - t^2)/t)`.

Answer 1

`x = cos^-1(4t^3 - 3t), y = tan^-1(sqrt(1 - t^2)/t)`
Put t = cosθ.
Thenθ = cos^–1t.
∴ x = cos^–1(4cos³θ  – 3cosθ ),

y = `tan^-1(sqrt(1 - cos^2θ)/cosθ)`
∴  x = `cos^-1(cos3θ),y = tan^-1((sinθ)/(cosθ)) = tan^-1(tanθ)`
∴ x = 3θ and y = θ
∴ x = 3cos^–1t and y = cos^–1t
Differentiating x and y w.r.t. x, we get
`"dx"/"dt" = 3"d"/"dt"(cos^-1)`

= `3 xx (-1)/sqrt(1 - t^2)`

= `(-3)/sqrt(1 - t^2)`
and
`"dy"/"dt" = "d"/"dt"(cos^-1t)`

= `(-1)/sqrt(1 - t^2)`

∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt")`

= `(((-1)/(sqrt(1 - t^2))))/(((-3)/(sqrt(1 - t^2)))`

= `(1)/(3)`.
Alternative Method :
x = cos^–1 (4t³ – 3t), t = `tan^-1(sqrt(1 - t^2)/t)`
Put t = cosθ.
Then x = cos^–1(4cos³θ – 3cosθ),

y = `tan^-1(sqrt(1- cos^2θ)/cosθ)`
∴ x = cos^–1 (cos3θ), y = `tan^-1((sinθ)/(cosθ)) = tan^-1(tanθ)`
∴ x = 3θ, y = θ
∴ x = 3y
∴ y = `(1)/(3)x`
∴ `"dy"/"dx" = (1)/(3)"d"/"dx"(x)`

= `(1)/(3) xx 1`

= `(1)/(3)`.